Chapter 11, Problem 3SR

It is often useful for companies to know who their customers are and how they became customers. A credit card company is interested in whether the owner of the card applied for the card on his or her own or was contacted by a telemarketer. The company obtained the following sample information regarding end-of-the-month balances for the two groups.

Is it reasonable to conclude the mean balance is larger for the credit card holders that were contacted by telemarketers than for those who applied on their own for the card? Assume the population standard deviations are not the same. Use the .05 significance level.

1. (a) State the null hypothesis and the alternate hypothesis.
2. (b) How many degrees of freedom are there?
3. (c) What is the decision rule?
4. (d) What is the value of the test statistic?
5. (e) What is your decision regarding the null hypothesis?
6. (f) Interpret the result.

To determine

State the null and the alternative hypothesis.

Answer to Problem 3SR

The null hypothesis is $H_0:{\mu }_1\ge {\mu }_2$

The alternative hypothesis is $H_1:{\mu }_1<{\mu }_2$

Explanation of Solution

In this context, ${\mu }_1$ denotes the mean balance for the credit card holders contacted by telemarketers and ${\mu }_2$ denotes the mean balance for those who applied on their won for the card.

The hypotheses are given below:

In this context,

Null hypothesis:

$H_0:{\mu }_1\ge {\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1<{\mu }_2$

Thus, the null hypothesis is $H_0:{\mu }_1\ge {\mu }_2$ and alternative hypothesis is $H_1:{\mu }_1<{\mu }_2$.

To determine

Find the number of degrees of freedom.

Answer to Problem 3SR

Thus, the number of degrees of freedom is 8.

Explanation of Solution

Calculation:

For t distribution with the two independent random samples, the degrees of freedom is as follows:

$df=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}$

Substitute $s_1$ as 356, $s_2$ as 857, $n_1$ as 10, and $n_2$ as 8 in the above formula as follows:

$\begin{array}{c}df=\frac{{\left(\frac{{356}^2}{10}+\frac{{857}^2}{8}\right)}^2}{\frac{{\left(\frac{{356}^2}{10}\right)}^2}{8-1}+\frac{{\left(\frac{{857}^2}{8}\right)}^2}{10-1}}\\ =\frac{{\left(12,673.6+91,806.125\right)}^2}{\frac{{\left(12,673.6\right)}^2}{9}+\frac{{\left(91,806.125\right)}^2}{7}}\\ =\frac{\text{10,916,012,936}\text{.075625}}{\text{17846681}\text{.88}+\text{1204052083}\text{.93}}\\ \approx 8\end{array}$

Thus, the number of degrees of freedom is 8.

To determine

Determine the decision rule.

Explanation of Solution

Step-by-step procedure to obtain the critical value using MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 8.
• Click the Shaded Area tab.
• Choose Probability and left Tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the critical value is –1.860.

The decision rule is as follows:

If $t<-1.860$, then reject the null hypothesis $H_0$.

To determine

Find the value of test statistic.

Answer to Problem 3SR

Thus, the value of the test statistic is –1.234.

Explanation of Solution

Calculation:

Test statistic:

The test statistics is obtained as follows:

$\begin{array}{c}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}}\\ =\frac{1,568-1,967}{\sqrt{\left(\frac{{356}^2}{10}+\frac{{857}^2}{8}\right)}}\\ =\frac{-399}{\sqrt{\text{104,479}\text{.725}}}\\ =\frac{-399}{323.233}\\ =-1.234\end{array}$

Thus, the test statistic is –1.234.

To determine

Determine the decision regarding $H_0$.

Answer to Problem 3SR

The decision is that fail to reject the null hypothesis $\left(H_0\right)$.

Explanation of Solution

Decision:

The critical value is –1.860 and the value of test statistic is –1.234.

The value of test statistic is greater than the critical value.

That is, $\text{Test statistic}\left(-1.234\right)>\text{Critical}\text{value}\left(-1.860\right)$.

From the decision rule, fail to reject the null hypothesis.

To determine

Determine the interpretation of the study.

Answer to Problem 3SR

The interpretation is that there is no evidence that the mean balance is larger for the credit card holders contacted by telemarketers than those who applied on their won for the card.

Explanation of Solution

Interpretation:

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the mean balance is larger for the credit card holders contacted by telemarketers than those who applied on their won for the card.